Let $f: R \rightarrow R$ be a differentiable function having $f(2)=6, f^{\prime}(2)=\left(\frac{1}{48}\right)$. Then $\lim _{x \rightarrow 2} \int_6^{f(x)} \frac{4 t^3}{x-2} d t$ equals

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$ \lim _{x ightarrow 2} \int_0^{f(x)} \frac{4 t^3}{x-2} d t $ Applying $L$ Hospital rule $ \begin{aligned} & \lim _{x ightarrow 2}\left[4 f(x)^2 f^{\prime}(x)ight]=4 f(2)^3 f^{\prime}(2) \ & =4 	imes 6^3 	imes \frac{1}{48}=18 . \end{aligned} $

Let $f: R \rightarrow R$ be a differentiable function having $f(2)=6,…

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